Author Topic: TUT0303 Quiz6  (Read 372 times)

Yiyang Huang

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TUT0303 Quiz6
« on: November 15, 2019, 02:00:40 PM »
a. Express the general solution of the given system of equations in terms of real-value functions.
$$
x^{\prime}=\left[\begin{array}{rr}{-2} & {1} \\ {1} & {-2}\end{array}\right] x
$$
$$
\begin{aligned} \operatorname{det}(A-\lambda I) &=\left|\begin{array}{cc}{-2-\lambda} & {1} \\ {1} & {-2-\lambda}\end{array}\right| \\ &=(-2-\lambda)(-2-\lambda)-1 \\ &=3+4 \lambda+\lambda^{2} \\ &=(\lambda+3)(\lambda+1) \end{aligned}
$$
$$
\lambda=-3, \lambda=-1
$$

when $\lambda=-3$, $(A-(-3) I) x=0$
$$
\left[\begin{array}{ll|l}{1} & {1} & {0} \\ {1} & {1} & {0}\end{array}\right] \sim\left[\begin{array}{ll|l}{1} & {1} & {0} \\ {0} & {0} & {0}\end{array}\right]
$$
$$
\begin{array}{c}{x_{2}=t, \quad x_{1}=-t} \\ {x=t\left[\begin{array}{c}{-1} \\ {1}\end{array}\right]}\end{array}
$$

when $\lambda=-1$, $(A-(-1) I) x=0$
$$
\left[\begin{array}{rr|r}{-1} & {1} & {0} \\ {1} & {-1} & {0}\end{array}\right] \sim\left[\begin{array}{rr|r}{1} & {-1} & {0} \\ {0} & {0} & {0}\end{array}\right]
$$
$$
\begin{aligned} x_{2} =t &\quad x_{1}=t \\ x=& t\left[\begin{array}{l}{1} \\ {1}\end{array}\right] \end{aligned}
$$

Hence, $y=c_{1} e^{-t}\left[\begin{array}{l}{1} \\ {1}\end{array}\right]+c_{2} e^{-3 t}\left[\begin{array}{c}{-1} \\ {1}\end{array}\right]$